We have seen how knowing the expectation enables us to set the odds of games. If you recall, the formula for calculating the expectation for any wager is :

(Probability of winning x the amount a wager would win) – (Probability of losing x the amount a wager would lose)

So, if you wanted to create your own casino game, you would have to go through the following process:

Calculating Probabilities of outcomes

Calculating the Expectation

Setting the Odds

From the previous posts, you would have already figured out 2 and 3.

How do we then get the probabilities of games? This is perhaps the most fundamental aspect of casino games. To do this, we would need to learn about combinatorial analysis.

Combinatorial analysis is about counting possibilities. This is essential when trying to determine the probabilities of events, such as a pair in Baccarat or a full house in poker. Knowing how to calculate these probabilities then allows you to derive expectation and from there, you can set the odds.

The 2 basic categories of combinatorial analysis are permutations and combinations.

Dr. James Tanton developed a brilliant method to calculate permutations and combinations. His use of labelling is highly effective, fast and easy to learn.

http://gdaymath.com/courses/permutations-and-combinations/

However, if you are keen to learn the slower and conventional method, read on…

Permutations and combinations can be further sub-divided into 2 categories, permutations or combinations with or without repeatable elements. Ready?

Permutations where you can repeat elements

The alphabet has 26 letters from A to Z. If you were to make a permutation of 3 REPEATABLE letters, you’d be able to repeat them, like AAA or BBA or BAB or ABB or XDX or DXX or XXD.

In a permutation, ABB would count as 1 outcome, just as BAB would count as 1 and BBA would count as 1. They contain the same letters, but in a different order.

So, the number of permutations you’d be able to make would be 26 x 26 x 26 = 17,576

A similar example would be 3 dice in 3 separate glasses. Let’s call them D1, D2 and D3. The dice are tumbled 3 times with the following results:

D1

D2

D3

1

2

2

2

1

2

2

2

1

Although the number 1 appears once and 2 appears twice in each permutation, they each count as 1 separate outcome by themselves. The total number of outcomes is then 3. So, for 3 dice, the total number of permutations would be = 6 x 6 x 6 = 216

Permutation where you can’t repeat

Now, if you were to make a permutation of 3 UNREPEATABLE letters, every time you used a letter, it could not be used again, like ABC, TRF or ZLD. This means that your pool of letters gets reduced each time you use one.

But remember, just like permutations with repeatable letters, permutations like ABC or CAB or BAC would each count as 1 possible outcome in themselves. They may contain the same letters, but their difference in order makes them each 1 distinct outcome.

The number of permutations you’d then be able to get for a 3 unrepeatable letter permutation is = 26 x 25 x 24 = 15,600

A similar example would be how many 52-card permutations you could make from a 52-card deck where each time a card is drawn, it is then discarded. The number of permutations for that would be = 52 x 51 x 50 x …x1

Permutations are generally used when we want to compute the total possible outcomes for all events.

Factorials

By now you would have noticed that there are certain formulas that multiply numbers in descending order, like 3 x 2 x 1 or 52 x 51 x 50 x 49…x1. This is called a factorial number and is symbolised in this fashion 3! – being 3 x 2 x 1. So, 6! would be 6 x 5 x 4 x 3 x 2 x 1.

Combinations where you can repeat elements

Combinations are sequences where we are only concerned with the appearance of certain elements. Computing combinations is used mainly for determining the categories of possible outcomes.

Let’s return to the 26-letter alphabet example. If we wanted a 3-letter combination from the alphabet, sequences like ABB, BBA or BAB would count as 1, just as XXY, XYX and YXX would also count as 1. The number of times the letters appear is what is important – how they are arranged is not.

Which, if you look at the table below, showing the possible combinations from a 3-dice roll, is accurate:

S/No

Combination

1

111

2

112

3

113

4

114

5

115

6

116

7

122

8

123

9

124

10

125

11

126

12

133

13

134

14

135

15

136

16

144

17

145

18

146

19

155

20

156

21

166

22

222

23

223

24

224

25

225

26

226

27

233

28

234

29

235

30

236

31

244

32

245

33

246

34

255

35

256

36

266

37

333

38

334

39

335

40

336

41

344

42

345

43

346

44

355

45

356

46

366

47

444

48

445

49

446

50

455

51

456

52

466

53

555

54

556

55

566

56

666

Remember that computing the combinations where elements can be repeated ALSO counts in combinations where elements can’t be repeated. They will all be counted. This is essential to remember when you are designing your own games.

Knowing how to count the possible combinations of events is helpful in planning the layout of a table game.

Combinations where you can’t repeat elements

A good example for this is when you want to determine how many 3-dice combinations there are when you cannot repeat numbers. In Sic-Bo, this is known as a 3-Single-Dice Combination. This counts 123, 321 and 231 as 1 possible outcome only.

The formula for this is

= number of numbers on 1 die! / number of dice! x (number of numbers on 1 die- number of dice)!

= 6! / 3! X (6-3)!

=6! / 3! X 3!

= 20

Which again, if you examine the table below, is accurate:

S/No

Combination

1

123

2

124

3

125

4

126

5

134

6

135

7

136

8

145

9

146

10

156

11

234

12

235

13

236

14

245

15

246

16

256

17

345

18

346

19

356

20

456

These are foundational principles in combinatorial analysis.

So, how do you count permutations and combinations in a practical way?

A great example is the calculation of how many 3-of-a-kinds are possible from a 52-card deck in a 5-card poker hand.

This will be deemed as a combination question as the order in which the cards appear is irrelevant – the casino pays if the hand contains a 3-of-a-kind, regardless of the order in which it appears in your hand.

Now, we know that the 52-card deck is really 13 numbers from Ace to King, with 4 copies in the form of suits, Hearts, Clubs, Diamonds and Spades.

So, following the labelling technique, we have 13 numbers in total. Any one of these could be the 3-of-a-kind, so we label 1 number as the 3-of-a-kind, 2 numbers as the remaining cards in the hand (remember, that these 2 numbers should not form a pair, or we’d get a Full House!) and the remaining 10 numbers as those not included in the hand.

We have to settle the suit portion. So, we’ll decide the suits of the 3-of-a-kind as well as the other 2 cards. We have 4 suits, of which the 3-of-a-kind must have 3. The other 2 cards will be of any suit as well.

Part 2 of the calculation looks like this:

= 4!/(3! X 1!) x 4!(1!/3!) x 4!(1!/3!) = 64

Finally, the total combinations containing a 3-of-a-kind is =

858 x 64 = 54,912.

To determine the probability, we divide the possible outcomes for a 3-of-a-kind by the total possible outcomes of 2,598,960

= 54,912/2,598,960 = 0.021128 or 2.1128%

What is the expectation of a 3-of-a-kind, then?

= 0.021128 – (1-0.021128) = 0.021128 – 0.978872 = -0.95774

And our break-even payment would be:

The losing probability / the winning probability

= 0.978872 /0.021128 = 46.33 (this is the point in terms of odds at which the expectation would be 0)

This is a huge edge for the house. Most poker games pay out 3:1 for a 3-of-a-kind! The expectation would be 0 if the house paid 46.33 times! So, if you are looking for an edge over the competition, perhaps it wouldn’t hurt to pay a little higher?

Of course, most poker games pay a jackpot for a straight flush or royal flush.

A royal flush has 4 combinations, with the following calculation:

Total outcomes = 52!/(5! X 47!) = 2,598,960

There are only 5 cards out of 13 that can be used for a royal flush (A,K,Q,J,10) = 5!/5! = 1

So, as long as the jackpot + wager payout for a royal flush does not exceed 649,739 times of the original wager, expectation should still be in favour of the house.

If the original back bet is $10 and assuming the payment for a royal flush is 100 times, then so long as the jackpot payout is LESS THAN 649,739 – 100 = 649,639 times the $10 wager or $6,496,390, the expectation should still be in the favour of the house. Of course, there’s no need to do this, but this establishes the upper limits of the maths.

Now, why is this even useful?

Knowing the total possible outcomes for an event allows us to calculate its probability.

Being able to do that then allows us to calculate the expectation of an event and to match payment odds according to the level of house edge desired.

We would also know how much further we’d be able to calibrate the odds in our payments

Now, the most important question

How much can I increase my payment odds before the expectation goes against the house?

The answer is simple = probability of losing / probability of winning

You can test this in a simple manner. Let’s have a single die roll. The probability of winning on number 1 is 1/6. The probability of losing is 5/6. Let’s assume a 1:1 payment.

If we divide 5/6 by 1/6, we get =0.833333 / 0.166667 = 5 (at this point, the expectation for the house would be 0)

Let’s test this out. Assuming now that we make our payment odds 5:1 for winning on number 1, our expectation would be = (0.166667 x 5) – (0.833333 x 1) = 0

As long as you keep your payment odds below this level, the expectation will always be in favour of the house.

Trust me, the players know this too.

The odds for many games have already been optimized to their limit – any higher and the advantage would go to the player!

Of, course the edge for the house has to be balanced with the ability to entice a player to wager in the first place. Understanding the mechanics underpinning your casino’s games goes a long way to maximising their profitability.

Who knows, you may even create your own best-selling game!

Appendix:

Sic-Bo Calculations

Wager

Combinations

Permutations

Specific Triples

6! / (1! X5!)

= 720 / 1 x 120

= 720 / 120

= 6 (6/6 numbers = 1 combination per number from 1 to 6.)

3! / (3!)

= 6 / 6

= 1 per combination

Specific Doubles

(this excludes specific triples)

6! / (1! X 1! X 4!)

= 720 / 24

= 30 ((30/6 numbers = 5 combinations per number from 1 to 6.)

3! / (2! X 1!)

= 6 / 2

= 3 per combination

2 Dice and Single Die Combination

6! / (1! X 1! X 4!)

= 720 / 24

= 30

3! / (2! X 1!)

= 6 /2

= 3 per combination

3 Single Die Combination

6! / (3! X 3!)

= 720 / 6 x 6

=720 / 36

=20

3! / (1! X 1! X 1!)

= 6 / 1

= 6 per combination

2 Dice Combination

6! / (2! X 4!)

= 720 / 2 x 24

= 720 / 48

= 15

Each combination would have 4 combinations where all 3 numbers are unique and 2 combinations where one of the numbers is a repeat, so: